# $bc$ CFT Energy-momentum tensor from Noether's theorem

Following Polchinski's book (String Theory 1), we have the $bc$ action:

$$S = \frac{1}{2 \pi}\int~d^2z ~b\bar \partial c,\tag{2.5.4}$$

where $b$ and $c$ have holomorphic weights $\lambda$ and $1- \lambda$.

From this action, it is said, that applying the Noether's theorem, we get the energy-momentum tensor:

$$T(z) = :(\partial b)c: - \lambda \partial(:bc:)\tag{2.5.11a}$$

With $\lambda=2$ (and so a holomorphic central charge $c=-26$), one has the correct Operator Product Expansions for Fadeev-Popov ghosts : $T(z) b(z)$ and $T(z)c(z)$, with holomorphic weights $(2,-1)$ so it is coherent, but, in fact, I am not able to apply the Noether's theorem/ Ward identities formalism used in chapter 2.3.

[So here, we may take $\lambda = 2$ for simplicity.]

(This question is related, while different, to a previous question :https://physics.stackexchange.com/a/69755/6316.)

So, the question, is, how get $(2.5.11a)$ from $(2.5.4)$?

[EDIT]

I was able to get this:

Starting from action:

$$S = \int d^2z g^{ab} b_{bc} \partial_a c^c \tag{1}$$

We make a variation:

$$\delta \sigma^d = \epsilon \rho(\sigma) v^d \tag{2}$$ The variation of the action, relatively to $\partial \rho$, is:

$$\delta S = (\epsilon g^{ab} b_{bc} v^d \partial_d c^c)~\partial_a \rho\tag{3}$$

This corresponds to a current:

$$j_a =i v^d(- b_{ac}\partial_d c^c) \tag{4}$$

The energy-momentum tensor is:

$$T_{ad} =- b_{ac}\partial_d c^c = (\partial_d b_{ac}) c^c- \partial_d(b_{ac} c^c)\tag{5}$$

Because only the holomorphic part is not null (from equations of movement?), that is $b=b_{zz}, c = c^z$ we have:

$$T(z) = T_{zz} = (\partial_z b) c- \partial_z (b c)\tag{6}$$

[EDIT 2]

There is a correct derivation in this reference (see formula 1.14), but it is using the standard derivation of the energy-momentum tensor from the action relatively to the metrics. So, it seems that it is the only way. Of course, at the end of the variation of the action, we may always write $\delta g_{\mu\nu}=\epsilon (v_\mu \partial_\nu \rho + v_\nu \partial_\mu \rho)$, and then get the correct current $(4)$ and the current energy-momentum tensor $(5)$. We have to take care to that $b_{zz}$ and $c^z$ do not depend explicitly on the metrics.

• Sorry, what's your question? – Luboš Motl Aug 10 '13 at 10:09
• @LubošMotl : I am not able to get 2.5.11a from 2.5.4 (in applying Noether's theorem) – Trimok Aug 10 '13 at 10:24
• Thanks, and what you are you getting from the procedure instead? Note that the second, $\lambda$-dependent term in 2.5.11a is the derivative of the ghost number and the coefficient of this term in the stress-energy tensor may only be determined if one considers the fields on a curved world sheet. So if your result is 2.5.11a with a more general or different value of $\lambda$, it's already OK for a flat world sheet. One may see that the $\lambda$ term has to be what it is because the role of $b,c$ must get reverted under $\lambda\leftrightarrow 1-\lambda$. – Luboš Motl Aug 10 '13 at 10:43
• @LubošMotl : Unfortunately, I am not able to apply the procedure, so I did not get anything. I am thinking to your comment. – Trimok Aug 10 '13 at 10:46
• @LubošMotl : I have made an EDIT to the question with what I was able to get , but the final result is wrong. – Trimok Aug 10 '13 at 14:20

You got just the wrong coefficient in front of $\partial_z(bc)$, you expected twice as high coefficient due to $\lambda=2$, right? But that's because since the very beginning, your Noether derivation was completely insensitive to the value of $\lambda$, so you got a random value of $\lambda$ by this derivation.

Your derivation is not sensitive to the right value of $\lambda$ because the initial action may be changed by any multiple of $\partial_z(bc)$ which is a "total derivative", and similarly for the antiholomorphic terms. Such total derivatives additions don't change the equations of motion but they change the coefficientin front of the $\partial(bc)$ term in your Noether-derived stress-energy tensor.

It's not hard to see that already eqn 2.5.4 is sloppy about this issue: it only differentiates $c$ but not $b$, by choice, although these two fields apparently play the same role and get "exactly" interchanged when $\lambda\leftrightarrow 1-\lambda$.

There exists the "right" coefficient in front of $\partial(bc)$ that one should add in the action that yields the correct Noether stress-energy tensor, and this choice could be perhaps "justified" as being special. But this is a waste of time. Finally, you want to derive a quantum stress-energy tensor (there is normal ordering etc. in it) and the Noether procedure is only OK to guess the right "classical limit" of the tensor without all the quantum subtleties.

The natural coefficient in front of $\partial(bc)$ in the stress-energy tensor cannot be canonically guessed by any of the classical methods because this term is automatically conserved, anyway: note that $\partial_z\partial_{\bar z}$ acting on a holomorphic or antiholomorphic function vanishes. The Noether method is a method to calculate a conserved current, so if there are many, it doesn't necessarily pick the right one. The right addition must be determined by demanding the right OPEs or some other "fully quantum" condition.

It's an old question though, I would write down my calculation for my own reference.

According to the textbook, $b$ and $c$ fields transform as tensors with weights $(\lambda,0)$ and $(1-\lambda,0)$, so we can get the following infinitesimal variations of conformal transformation. $$z'=z+\epsilon v(z)\\ \delta b\equiv b'(z)-b(z)=-\epsilon(v\partial b+\lambda\partial v b)\\ \delta c\equiv c'(z)-c(z)=-\epsilon(v\partial c+(1-\lambda)\partial v c)$$ Multiply $\delta b$ and $\delta c$ by $\rho(\sigma)$ and plug into the action, we get $$-\delta S=\frac{\epsilon}{2\pi}\int d^2z\left(\lambda (\bar\partial\rho) v\partial (bc)-(\bar\partial\rho)v(\partial b)c-\lambda (\partial\rho) v\bar\partial (bc)+(\partial\rho)v(\bar\partial b)c\right).$$ Compare this with Eq. 2.3.4 in the textbook, and note that $d^2z=2d^2\sigma$, $j^z=2j_{\bar z}$, $j^{\bar z}=2j_z$, we can get $$T(z)=\frac{j_z}{iv(z)}=(\partial b)c-\lambda\partial(bc)~~(\text{before normal ordering})\\ \tilde T(\bar z)=\frac{j_{\bar z}}{iv^*(\bar z)}=-(\bar\partial b)c+\lambda\bar\partial(bc)~~(\text{before normal ordering})$$ Now put on normal ordering and note that $:bc:$ is purely holomorphic, we get Eq. 2.5.11.

There is still a tricky problem here. The above calculation only considers the variation of the action and forgets about the integral measure. Since $b$ and $c$ fields have a nontrivial scaling here, I would expect there be a nontrivial Jacobian, but I don't know how to compute it.

I am adding my own computation to the mix since it took me a while to follow the Noether method of deriving eq. (13.17) in Blumenhagen, Lüst, Theisen. The setup is the same as in Polchinski.

The fields $$b$$ and $$c$$ have conformal weight $$\lambda$$ and $$1-\lambda$$ and their action is given by

$$S = \frac{1}{2\pi} \int d^2z \, b \bar \partial c$$

Recall that under conformal transformations $$z \rightarrow z+\epsilon(z)$$

$$b(z) \rightarrow \left(1 + \frac{\partial \epsilon}{\partial z} \right)^\lambda b(z+\epsilon(z)) = b(z) + \lambda \partial\epsilon(z)b(z) + \epsilon(z)\partial b(z) + \mathcal{O}(\epsilon^2)$$

and similarly for $$c(z)$$. The way to derive the Noether currents is to pretend that the parameter $$\epsilon(z)$$ depends on the coordinates $$z$$ and $$\bar z$$. I believe this point to be a bit subtle since I was assuming $$\bar \partial \epsilon = 0$$ using the fact that it is a conformal transformation which should obviously not depend on $$\bar z$$. Anyway, allowing such behavior, the variation of $$S$$ then is

\begin{align} \delta S &= \frac{1}{2\pi} \int d^2z \,[\lambda \partial \epsilon b + \epsilon \partial b]\bar \partial c + b\bar\partial [(1-\lambda)\partial \epsilon c + \epsilon \partial c] \\ &= \frac{1}{2\pi} \int d^2z \, \left(\lambda \epsilon \partial(b\bar\partial c) + \epsilon \partial b \bar \partial c + (1-\lambda)(b\bar \partial \partial \epsilon c + b\partial \epsilon \bar \partial c) \\+ b \bar \partial \epsilon \partial c + \epsilon b \bar \partial \partial c \right) ~. \end{align}

It is not hard to check that all terms cancel after integrating by parts as usual except for the two terms containing barred derivatives of $$\epsilon$$. One is left with the following expression

\begin{align} \delta S &= \frac{1}{2\pi} \int d^2z \, (1-\lambda)b \bar \partial \partial \epsilon c + b \bar \partial \epsilon \partial c \\ &= \frac{1}{2\pi} \int d^2z \, \epsilon \bar \partial[(1-\lambda) \partial (bc) - (b\partial c)] \\ &= \frac{1}{2\pi} \int d^2z \, \epsilon \bar \partial[-\lambda b \partial c + (1-\lambda) \partial b c] \\ &\cong \frac{1}{2\pi} \int d^2z \, \epsilon(z)\partial_{\bar z} T(z)~, \end{align}

where $$T(z)$$ now has the desired form after decorating it with normal-ordering symbols:

$$T(z) = - \lambda :b \partial c: + (1-\lambda) :(\partial b) c:$$

• You do an implicit integration by parts in the fourth to last line, $(1 - \lambda) b \overline{\partial} \partial \epsilon c \mapsto -(1 - \lambda) \overline{\partial} \epsilon \partial(bc)$. This only works if the boundary term is $0$, meaning you have to impose that $\overline{\partial}\epsilon = 0$ on the bounary of integration. There is nothing wrong with this, just thought I'd point it out. – user1379857 Oct 6 '19 at 18:03